Is #1 / ( x + 3 )^2# strictly greater than # 2/(x-3)#?

1 Answer
Oct 26, 2015

Answer:

No (not always, at least).

Explanation:

If it was, then the answer to "For which values of #x# the following disequality holds?"

#1/(x+3)^2 - 2/(x-3) >0#

Should be "Always". Let's check it. Make the fractions have a common denominator, and you'll end up with

#(x-3)/((x+3)^2(x-3)) - (2(x+3)^2)/((x+3)^2(x-3)) >0#.

Now sum the numerator:

#(x-3)-2(x+3)^2=x-3-2(x^2+6x+9)=#

#=x-3-2x^2-12x-18=#

#=-2x^2-11x-21 =-(2x^2+11x+21)#

Since #2x^2+11x+21# has a negative discriminant, it has no solution, and so it is always positive. So, the fraction becomes

#- (2x^2+11x+21) / ((x+3)^2(x-3))#

and we know that #2x^2+11x+21# and #(x+3)^2# are always positive (the latter being a square). So, the sign of the whole fraction depends on the sign of #x-3#, which obviously is positive after #3# and negative before. Since there is a minus before the whole fraction, we have that the expression is positive only before #3#, and so the answer is no:

#1/(x+3)^2# is not always greater than #2/(x-3)#, but only before #3#.